Problem: The function $f$ is not defined for $x = 0,$ but for all non-zero real numbers $x,$
\[f(x) + 2f \left( \frac{1}{x} \right) = 3x.\]Find the real solutions to $f(x) = f(-x).$  Enter the real solutions, separated by commas.
We are given
\[f(x) + 2f \left( \frac{1}{x} \right) = 3x.\]Replacing $x$ with $\frac{1}{x},$ we get
\[f \left( \frac{1}{x} \right) + 2f(x) = \frac{3}{x}.\]We can view these equations as a system in $f(x)$ and $f \left( \frac{1}{x} \right).$  Solving for $f(x),$ we find
\[f(x) = \frac{2 - x^2}{x}.\]Then the equation $f(x) = f(-x)$ becomes
\[\frac{2 - x^2}{x} = \frac{2 - x^2}{-x}.\]Then $2 - x^2 = x^2 - 2,$ so $x^2 = 2.$  The solutions are $\boxed{\sqrt{2},-\sqrt{2}}.$